Question: A particle moves in the $xy$ -plane so that at any time $t\geq 0$ its coordinates are $x=t^3+4t$ and $y=t^2+8t$. What is the magnitude of the particle's velocity vector at $t=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $26$ (Choice B) B $4\sqrt{6}$ (Choice C) C $11$ (Choice D) D $\sqrt{149}$
Background When solving motion problems, it's important to remember the relationship between the position vector $(x,y)$, the velocity vector $\vec{v}(t)$, and the acceleration vector $\vec{a}(t)$ of the moving particle: $\vec{v}(t)=\left(\dfrac{dx}{dt},\dfrac{dy}{dt}\right)$ $\vec{a}(t)=\dfrac{d}{dt}\vec{v}(t)=\left(\dfrac{d^2x}{dt^2},\dfrac{d^2y}{dt^2}\right)$ Setting up the math We are given that the particle's coordinates are $x=t^3+4t$ and $y=t^2+8t$, which means its position vector is $(t^3+4t,t^2+8t)$. We are asked to find the magnitude of the particle's velocity vector at $t=1$. In other words, we need to find $||\vec{v}(1)||$. Finding $\vec{v}(t)$ $\begin{aligned} \vec{v}(t)&=\left(\dfrac{d}{dt}(t^3+4t),\dfrac{d}{dt}(t^2+8t)\right) \\\\ &=(3t^2+4,2t+8) \end{aligned}$ Finding $\vec{v}(1)$ $\begin{aligned} \vec{v}({1})&=(3({1})^2+4,2({1})+8) \\\\ &=(3+4,2+8) \\\\ &=(7,10) \end{aligned}$ Finding $||\vec{v}(1)||$ $\begin{aligned} ||\vec{v}(1)||&=||(C{7},{10})|| \\\\ &=\sqrt{(C{7})^2+({10})^2} \\\\ &=\sqrt{149} \end{aligned}$ In conclusion, the magnitude of the particle's velocity vector at $t=1$ is $\sqrt{149}$.